Describe how you would prepare approximately 2 l of 0.050 0 m boric acid, b(oh)3.

The given concentration of boric acid = 0.0500 M

Required volume of the solution = 2 L

Molarity is the moles of solute present per liter solution. So 0.0500 M boric acid has 0.0500 mol boric acid present in 1 L solution.

Calculating the moles of 0.0500 M boric acid present in 2 L solution:

Converting moles of boric acid to mass:

Therefore, 6.183 g boric acid when dissolved and made up to 2 L with distilled water gives 0.0500 M solution.

8.75 g

We are given;

Mass of Os-182 sample = 10.0 g

Number of half lives = 3

Half life = 21.5 hours

We need to determine the mass that decayed after 3 half lives

Using the formula;

Remaining mass = Initial mass × (1/2)^n , where n is the number of half lives

Therefore;

Mass remaining = 10.0 g × (1/2)^3

                          =1.25 g

Thus, to get the mass that decayed;

Mass that decayed = Initial mass - remaining mass

                                = 10.0 g - 1.25 g

                                = 8.75 g

Hence, the mass of Os-182 that decayed is 8.75 g